The matrix exponential function of an m×m matrix M is defined as

Exp(M) = I/0! + M/1! + M²/2! + … + Mⁿ/n! + …

where I is the m×m identity matrix. The series converges for all M.

If M is purely imaginary, such as iQ where the elements of Q are all real, then

Exp(iQ) = Cos(Q) + i·Sin(Q)

The topic pursued here is the proof that the matrix cosine and sine functions satisfy the common trigonometric identities, such as cos²(φ)+sin²(φ)=1.

First note

Lemma 0:

If O is any square matrix of all zeroes then

Exp(O) = I

Putting O into the defining series results in all terms except the first being equal to O.

Then note that the complex conjugate of Exp(iQ) is Exp(−iQ). Thus

Exp(iQ)Exp(−iQ) = (Cos(Q)+i·Sin(Q))·(Cos(Q)−i·Sin(Q))
= Cos²(Q) + Sin²(Q)
But Exp(iQ)Exp(−iQ) =Exp(i(Q−Q) = Exp(O) = I

Thus

Lemma 1:

Cos²(Q) + Sin²(Q) = I

Lemmas 2 and 3:

Cos(Q) = [Exp(iQ) + Exp(−iQ)]/2
 
Sin(Q) = [Exp(iQ) − Exp(−iQ)]/(2i)

In dealing with the product of exponential care must be taken because

exp(A)·exp(B) = exp(A+B)
only if
AB = BA

In the above this condition is trivially satisfied.

Now consider [exp(A)]ⁿ for n being a nonnegative integer. Then

Lemma 5:

[exp(A)]ⁿ = exp(nA)

Then

Lemma 6:

Cos²(Q) = [Cos(2Q) +I]/2

Since Cos(Q)=[Exp(iQ) + Exp(−iQ)]/2

Cos²(Q) = {[Exp(iQ)]² + 2Exp(iQ)·Exp(−iQ) + [Exp(−iQ)]²}/4
which reduces to
Cos²(Q) = [Exp(2iQ) + Exp(−2iQ) + 2I]/4
and further to
Cos²(Q) = [Cos(2Q) + I]/2

Likewise

Lemma 7:

Sin²(Q) = [I − Cos(2Q)]/2

Since Sin(Q)=[Exp(iQ) − Exp(−iQ)]/2i

Sin²(Q) = {[Exp(iQ)]² − 2Exp(iQ)·Exp(−iQ) + [Exp(−iQ)]²}/(−4)
which reduces to
Sin²(Q) = [Exp(2iQ) + Exp(−2iQ) − 2I]/(−4)
and further to
Sin²(Q) = [I −Cos(2Q)]/2

Consider Sin(Q)Cos(Q). This is

Sin(Q)Cos(Q) = {[Exp(iQ) − Exp(−iQ)]/2i}{[Exp(iQ) + Exp(−iQ)]/2}
which evaluates to
Sin(Q)Cos(Q) = {[Exp(iQ)]² − [Exp(−iQ)]²}/(4i)
which reduces to
Sin(Q)Cos(Q) = {[Exp(i2Q)] − [Exp(−i2Q)]}/(4i)
and further to
Sin(Q)Cos(Q) = Sin(2Q)/2

Thus

Lemma 8:

Sin(Q)Cos(Q) = Sin(2Q)/2 = [I − Cos(2Q)]/2

And likewise

Lemma 8a:

Cos(Q)Sin(Q) = Sin(2Q)/2 = [I − Cos(2Q)]/2

And furthermore

Lemma 9:

Sin(Q)Cos(Q) = Cos(Q)Sin(Q)

Now define

Tan(Q) = [Cos(Q)]⁻¹Sin(Q)

and then consider

Cos²(Q)[I + Tan²(Q)]

This reduces successively to

Cos²(Q)[I + Tan²(Q)] = Cos²(Q) + Cos(Q)Sin(Q)[Cos⁻¹(Q)Sin(Q)]
= Cos²(Q) + Sin(Q)Cos(Q)Cos⁻¹(Q)Sin(Q) = Cos²(Q) + Sin²(Q) = I

Thus

Lemma 10:

I + Tan²(Q) = [Cos²(Q)]⁻¹ = [Cos(Q)⁻¹]²

where Cos(Q)⁻¹ could be denoted as Sec(Q).

Similarly if Cot(Q)=Sin⁻¹(Q)Cos(Q)

Lemma 11:

I + Cot²(Q) = [Sin²(Q)]⁻¹ = [Sin(Q)⁻¹]²

where Sin(Q)⁻¹ could be denoted as Csc(Q).

Finally to make things complete

Lemma 12:

Exp(iπI) = −I

(To be continued.)